In Figure 32.20, the capacitor is initially charged when switch S1 is open and S2 is closed. Switch S2 is then opened, removing the battery from the circuit, and the capacitor remains charged. Switch S1 is then closed, so that the capacitor is connected directly across the inductor.

Question

In Figure 32.20, the capacitor is initially charged when switch $S_{1} \text { is open and } S_{2} \text { is closed. Switch } S_{2}$ is then opened, removing the battery from the circuit, and the capacitor remains charged. Switch $S _{1}$ is then closed, so that the capacitor is connected directly across the inductor.

(A) Find the frequency of oscillation of the circuit.

(B) What are the maximum values of charge on the capacitor and current in the circuit?

(C) Determine the charge and current as functions of time.

(A) Find the frequency of oscillation of the circuit.

(B) What are the maximum values of charge on the capacitor and current in the circuit?

(C) Determine the charge and current as functions of time.

Accepted Answer

Using Equation 32.22 gives for the frequency

[latex] \omega=\frac{1}{\sqrt{L C}} [/latex] (32.22)

[latex] f=\frac{\omega}{2 \pi}=\frac{1}{2 \pi \sqrt{L C}} [/latex].

[latex] =\frac{1}{2 \pi\left[\left(2.81 imes 10^{-3} H ight)\left(9.00 imes 10^{-12} F ight) ight]^{1 / 2}} [/latex].

[latex] =1.00 imes 10^{6} Hz [/latex].

(B) What are the maximum values of charge on the capacitor and current in the circuit?

The initial charge on the capacitor equals the maximum charge, and because [latex] C=Q / \Delta V [/latex], we have

[latex] Q_{\max }=C \Delta V=\left(9.00 imes 10^{-12} F ight)(12.0 V ) [/latex].

[latex] =1.08 imes 10^{-10} C [/latex].

From Equation 32.25, we can see how the maximum current is related to the maximum charge:

[latex] I=-\omega Q_{\max } \sin \omega t=-I_{\max } \sin \omega t [/latex] (32.25).

[latex] I_{\max }=\omega Q_{\max }=2 \pi f Q_{\max } [/latex].

[latex]=\left(2 \pi imes 10^{6} s ^{-1} ight)\left(1.08 imes 10^{-10} C ight) [/latex].

[latex] I_{\max }=6.79 imes 10^{-4} A [/latex].

(C) Determine the charge and current as functions of time.

Equations 32.24 and 32.25 give the following expressions for the time variation of Q and I :

[latex] Q=Q_{\max } \cos \omega t [/latex] (32.24)

[latex] I=-\omega Q_{\max } \sin \omega t=-I_{\max } \sin \omega t [/latex] (32.25).

[latex] Q=Q_{\max } \cos \omega t [/latex].

[latex] =\left(1.08 imes 10^{-10} C ight) \cos \left[\left(2 \pi imes 10^{6} s ^{-1} ight) t ight] [/latex].

[latex] I=-I_{\max } \sin \omega t [/latex].

[latex] =\left(-6.79 imes 10^{-4} A ight) \sin \left[\left(2 \pi imes 10^{6} s ^{-1} ight) t ight] [/latex].

Question 32.7: In Figure 32.20, the capacitor is initially charged when swi...

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